When 78.0 g of aluminum hydroxide, Al(OH)₃ (molar mass = 78.0 g/mol), reacts with 49.0 g of sulfuric acid, H₂SO₄ (molar mass = 98.1 g/mol), what mass of water is produced? The equation is:

\[ 2\text{Al(OH)}_{3(s)} + 3\text{H}_2\text{SO}_{4(aq)} \rightarrow \text{Al}_2(\text{SO}_4)_{3(aq)} + 6\text{H}_2\text{O}_{(l)} \]

A. 18.0 g
B. 36.0 g
C. 54.0 g
D. 127 g

To determine the mass of water produced when 78.0 g of aluminum hydroxide reacts with 49.0 g of sulfuric acid, you need to first calculate the limiting reactant. The limiting reactant is the reactant that is completely consumed and determines the amount of product formed. In this case, you compare the number of moles of Al(OH)3 to the number of moles of H2SO4. The reactant with the smaller number of moles is the limiting reactant. The molar mass of water (H2O) is 18.0 g/mol. Using the balanced equation, you can calculate the mass of water produced by multiplying the number of moles of water by its molar mass.

Molar mass of Al(OH)3: 78.0 g/mol

Molar mass of H2SO4: 98.1 g/mol

Number of moles of Al(OH)3: 78.0 g / 78.0 g/mol = 1.00 mol

Number of moles of H2SO4: 49.0 g / 98.1 g/mol = 0.499 mol

Since there are 2 moles of Al(OH)3 for every 3 moles of H2SO4, the number of moles of H2SO4 required to react with 1.00 mole of Al(OH)3 is:

0.499 mol / 3 mol x 2 mol = 0.332 mol

Therefore, H2SO4 is the limiting reactant as it requires 0.332 moles to react completely, whereas Al(OH)3 only provides 0.333 moles.

To calculate the mass of water produced, we can use the molar ratio from the balanced equation:

6 moles of H2O are produced for every 2 moles of Al(OH)3 reacted.

Number of moles of H2O produced:

0.333 mol Al(OH)3 × (6 mol H2O / 2 mol Al(OH)3) = 1.00 mol H2O

Mass of water produced:

1.00 mol H2O × 18.0 g/mol = 18.0 g

Therefore, the mass of water produced is 18.0 g.

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